Formation of a molecular species

1. Formation of a molecular species • It is the same as precipitates or gases except a liquid is formed. • Acid base neutralization reactions will produce water. • NaOH + HNO3 H2O (l) + NaNO3(aq)

2. Strong acids

3. Strong Bases these make a lightning bolt on the periodic table!

4. Strong acids and bases • Strong acids and bases are not at equilibrium, there is no reverse reaction. • Strong acids and bases will never be formed in a net ionic equation. • All other acids/bases can be formed, and will be formed by reacting the appropriate ion with a strong acid/base. • *Most other bases are insoluble

5. Examples • Calcium hydroxide reacts with chloric acid • Hydrochloric acid reacts with calcium nitrite • Nitric acid reacts with sodium chlorite • Sodium chloride is mixed with sulfuric acid

6. Chapter 10 Aqueous Solutions and Ionic Equations • This chapter has already been covered • *Only dissociate soluble ionic compounds • Molecular equation • Na2S + CrCl2 CrS + NaCl • Full Ionic Equation • 2Na++S2-+Cr2++2Cl- CrS+Na++Cl- • Net Ionic Equation • Cr2++S2- CrS

7. Chapter 11 Redox Reactions • Redox or oxidation-reduction reactions are reactions that involve a transfer of electrons. • Oxidation is the loss of electrons. • Reduction is the gain of electrons. • (think of the charge, OIL RIG) • 4 K + O2→ 4 K+ + 2 O2- • Potassium get oxidized, oxygen get reduced

8. Using oxidation states • In the reaction… • 2 Na +2 H2O →2 NaOH + H2 • 0 +1 -2 +1 -2 +1 0 • Note the changes • Sodium went from 0 to 1 • 2 of the hydrogen atoms went from +1 to 0 (the other two were unchanged)

9. Breaking into two half reactions • Sodium must have lost 2 electrons • 2 Na → 2Na+ + 2 e- • And Hydrogen gained two electrons • 2 H2O +2 e-→ 2 OH- + H2 • Sodium is oxidized, hydrogen is reduced in this reaction • Oxidation is an increase in oxidation state • Reduction is a decrease in oxidation state

10. Balancing Redox Equations • by Half Reactions Method or oxidation state method • The book does not separate these into half reactions, although it adds another step I think it makes it easier

11. Half reactions • Ce4+ + Sn2+→ Ce3+ + Sn4+ • Half reactions • Ce4+ + e-→ Ce3+ • Sn2+→ 2e- + Sn4+ • Electrons lost must equal electrons gained! • 2Ce4+ +2 e-→2 Ce3+ • Merge the two half reactions • 2 Ce4+ + Sn2+→ 2 Ce3+ + Sn4+

12. Redox reactions in acidic solutions • It will be noted in the problem • Balance all elements except hydrogen and oxygen. • Balance oxygen by adding H2O (which is always prevalent in an acidic solution) • Balance hydrogen by adding H+ • Then balance the charge adding electrons and proceed as normal.

13. Example • In an acidic solution • Cr2O72- + Cl-→ Cr3+ + Cl2 • Half reactions • Cr2O72-→ Cr3+ • Cl-→ Cl2

14. Reduction side • Cr2O72-→ Cr3+ • Cr2O72-→ 2 Cr3+ • Cr2O72-→ 2 Cr3+ + 7 H2O • Cr2O72- + 14 H+→ 2 Cr3+ + 7 H2O • Cr2O72- + 14 H++ 6 e-→2Cr3++7 H2O

15. Oxidation side • Cl-→ Cl2 • 2 Cl-→ Cl2 • 2 Cl-→ Cl2 + 2 e- • I have to equal 6 e- so multiply by 3 • 6 Cl-→ 3 Cl2 + 6 e-

16. Combine my half reactions • Cr2O72- + 14 H++ 6 e- → 2 Cr3+ + 7 H2O • 6 Cl-→ 3 Cl2+ 6 e- • And you get • Cr2O72-+14 H++6Cl-→2Cr3++3 Cl2+7H2O • The electrons cancel out .

17. Example • In an acidic solution • MnO4- + H2O2→ Mn2+ + O2 • Half reactions • MnO4- → Mn2+ • H2O2→ O2

18. Top Equation • MnO4- → Mn2+ • MnO4- → Mn2+ + 4 H2O • MnO4- + 8 H+→ Mn2+ + 4 H2O • MnO4- + 8 H+→ Mn2+ + 4 H2O • MnO4- + 8 H++ 5 e-→ Mn2+ + 4 H2O

19. Bottom Equation • H2O2→ O2 • H2O2→ O2 + 2 H+ • H2O2→ O2 + 2 H+ + 2 e- • I need to equal 5 e- so… • That won’t work… • 2MnO4- + 16 H++ 10 e-→ 2 Mn2+ + 8 H2O • 5 H2O2→ 5 O2 + 10 H+ + 10 e-

20. Add them together • 2MnO4- + 16 H++ 10 e-→ 2 Mn2+ + 8 H2O • 5 H2O2→ 5 O2 + 10 H+ + 10 e- • And you get • 2 MnO4- + 6 H++ 5 H2O2 → 2 Mn2+ + 5 O2 + 8 H2O • Notice the H+ canceled out as well.

21. Balancing Redox Equations in a basic solution • Look for the words basic or alkaline • Follow all rules for an acidic solution. • After you have completed the acidic reaction add OH- to each side to neutralize any H+. • Combine OH- and H+ to make H2O. • Cancel out any extra waters from both sides of the equation.

22. Example • We will use the same equation as before • In a basic solution • MnO4- + H2O2→ Mn2+ + O2 • 2 MnO4- + 6 H++ 5 H2O2 → 2 Mn2+ + 5 O2 + 8 H2O

23. Basic solution • Since this is a basic solution we can’t have excess H+. • We will add OH- to each side to neutralize all H+ • 2 MnO4- + 6 H++ 5 H2O2 + 6OH- →2 Mn2+ +5 O2 +8 H2O + 6OH- • We added 6 OH- because there were 6H+

24. Cont. • H+ + OH- → H2O • Combine the hydroxide and hydrogen on the reactant side to make water • 2 MnO4- + 6 H2O + 5 H2O2 → 2 Mn2++5 O2+ 8 H2O + 6OH- • Cancel out waters on both sides 2 MnO4- + 5 H2O2 →2 Mn2+ + 5 O2 +2 H2O +6OH-

25. Another example • In a basic solution • MnO4 − + SO32-→MnO4 2− + SO42- • Half reactions • MnO4 − → MnO4 2− • SO32-→ SO42-

26. Half reactions • MnO4 − → MnO4 2− • MnO4 - + e- → MnO4 2− • SO32-→ SO42- • H2O + SO32-→ SO42- • H2O + SO32-→ SO42- + 2 H+ • H2O + SO32-→ SO42- + 2 H+ +2e- • Double the top reaction

27. 2 MnO4 - + 2 e- → 2 MnO4 2− • H2O + SO32-→ SO42- + 2 H+ +2e- • Combine them • 2 MnO4 - + H2O + SO32- → 2 MnO4 2− +SO42- + 2 H+ • Add OH- 2 MnO4 - + H2O + SO32- + 2 OH- → 2 MnO4 2−+SO42- +2 H++2 OH-

28. 2 MnO4 - + H2O + SO32- + 2 OH- → 2 MnO4 2− +SO42- + 2 H2O • finishing • 2 MnO4 - + SO32- + 2 OH- → 2 MnO4 2− +SO42- + H2O